What happens if you try to compile and run this program?

What happens if you try to compile and run this program?

#include <stdio.h>

int main (int argc, char *argv[]) {

char *t = "abcdefgh";

char *p = t + 2;

int i;

p++;

p++;

printf("%d ", p[2] – p[-1]);

C++ Institute – CLA-11-03

return 0;

}

Choose the right answer:
A . The program outputs 3
B . The program outputs 4
C . Execution fails
D . The program outputs 2
E . Compilation fails

View Answer

Answer: A

Explanation:

The program outputs 3 because the expression p[2] – p[-1] evaluates to 3 using the pointer arithmetic rules in

C. The pointer t points to the first element of the string literal "abcdefgh", which is stored in a read-only memory location. The pointer p is initialized to t + 2, which means it points to the third element of the string, which is ‘c’. Then, p is incremented twice, so it points to the fifth ele-ment of the string, which is ‘e’. The subscript operator [] is equivalent to adding an offset to the pointer and dereferencing it, so p[2] is the same as *(p + 2), which is ‘g’, and p[-1] is the same as *(p – 1), which is ‘d’. The printf function then prints the difference between the ASCII values of ‘g’ and ‘d’, which is 103 – 100 = 3, as a decimal integer using the %d format specifier.

References = CLA C C Certified Associate Programmer Certification, C Essentials 2 – (Intermediate), C Pointers, C Strings